# Proof that a finite separable extension has only finite many intermediate fields

Let $E/F$ be finite separable extension. Is there any proof of the fact that there are only finitely many intermediate fields without using primitive element theorem or fundamental theorem of Galois theory?

• What's the motivation for not wanting to use either of those results? Considering the primitive element theorem is often stated as a finite extension has a primitive element if and only if there are a finite number of intermediate fields. Commented Feb 7, 2013 at 14:48
• If one cannot use galois-theory, then why tag this galois-theory? Commented Feb 7, 2013 at 15:00
• Because I thought it is related to galois theory or can be used in galois theory. Commented Feb 7, 2013 at 15:02
• Indeed. Thanks for the clarification. Commented Feb 7, 2013 at 15:11
• @Mohan: Lang in his Algebra shows directly that a finite separable extension has a primitive element, and thus your result follows using the (stronger form of) The Primitive Element Theorem but no Galois Theory. (This argument is replicated in $\S 8$ of my field theory notes: math.uga.edu/~pete/FieldTheory.pdf.) I think this is not what you are looking for. Could you amplify on why you want to avoid the Primitive Element Theorem and Galois Theory? Commented Mar 3, 2013 at 17:57

1) Let $K$ be a field and $n\in \mathbb{N}$. Then $K^n$ has only finite many commutative unital subalgebras. The exact number is $B(n)$ - the $n$th Bell number.
2) Let $(K;L)$ be a separable extension. Then $L$ is separable as $K$-algebra. Thus, by tensoring a suitable splitting field $T$ for $L$ we get that $L\otimes T$ is isomorphic to $T^n$ for a suitable $n$. Hence - using 1) - $L\otimes T$ has only finite many commutative unital subalgebras.
3) The function $A\mapsto A\otimes L$ is injective defined on the commutative unital subalgebras of $L$.
4) Any commutative unital subalgebra of $L$ is a field, too.
• I don't think that 3) is correct. Let $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2}, \sqrt{3})$, then $\mathbb{Q}(\sqrt{2}) \otimes \mathbb{Q}(\sqrt{2}, \sqrt{3}) \cong \mathbb{Q}(\sqrt{3}) \otimes \mathbb{Q}(\sqrt{2}, \sqrt{3})$.